JEE Main 2026 — Physics Mechanics

2026mcqhard

A wedge $Y$ with mass of $10$ kg and all frictionless surfaces and the inclined surface making $37°$ with horizontal. A block $X$ with mass $2$ kg is placed at the highest point of the wedge as shown in figure is at rest. At $t = 0$ wedge $(Y)$ is pulled toward right with constant force $(f)$ of $24$ N. Taking the block $X$ at rest at $t = 0$ , the time taken by it to slide down $8.8$ m on the slope, while $Y$ is on the move, is _____ s.

(take $\tan(37°) = 3/4$ and $g = 10$ m/s $^2$ )

JEE Main 2026 Physics, Laws of Motion — question figure

Official previous-year question

Held on 5 Apr 2026 · Verified 6 Jul 2026.

Options

A
$2$
B
$4$
C
$\sqrt{2}$
D
$2\sqrt{2}$

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Solution

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):

$A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2$

Working in the non-inertial frame of the wedge, the block experiences a pseudo force $mA$ directed opposite to the wedge's acceleration. The components along the incline are:

Component of gravity down the incline: $g\sin 37^\circ$

Component of pseudo force up the incline: $A\cos 37^\circ$

Net relative acceleration of the block down the incline:

$a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ$

Using $\sin 37^\circ = 0.6$ and $\cos 37^\circ = 0.8$ :

$a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2$

Time taken to slide a distance of $8.8\text{ m}$ along the incline, starting from rest:

$s = \dfrac{1}{2} a_{\text{rel}}\, t^2$

$8.8 = \dfrac{1}{2}(4.4)\,t^2$

$8.8 = 2.2\,t^2$

$t^2 = 4 \Rightarrow t = 2\text{ s}$

Hence, the required time is $2\text{ s}$ .

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Solution

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):

$A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2$

Working in the non-inertial frame of the wedge, the block experiences a pseudo force $mA$ directed opposite to the wedge's acceleration. The components along the incline are:

Component of gravity down the incline: $g\sin 37^\circ$

Component of pseudo force up the incline: $A\cos 37^\circ$

Net relative acceleration of the block down the incline:

$a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ$

Using $\sin 37^\circ = 0.6$ and $\cos 37^\circ = 0.8$ :

$a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2$

Time taken to slide a distance of $8.8\text{ m}$ along the incline, starting from rest:

$s = \dfrac{1}{2} a_{\text{rel}}\, t^2$

$8.8 = \dfrac{1}{2}(4.4)\,t^2$

$8.8 = 2.2\,t^2$

$t^2 = 4 \Rightarrow t = 2\text{ s}$

Hence, the required time is $2\text{ s}$ .

JEE Main 2026 — Physics Mechanics

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JEE Main 2026 — Physics Mechanics

Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics